The first-null bandwidth, or main-lobe width, of baseband signals is related to their symbol rate and modulation scheme. For M-PSK and M-QAM signals, it’s commonly calculated based on the Nyquist theorem, which states that the minimum bandwidth for zero inter-symbol interference is the symbol rate. However, for practical signals (e.g., raised-cosine filtering), the bandwidth may vary slightly depending on factors such as excess bandwidth.

Given:

• Symbol rate $f_s = 8.3 \, \text{k sym/s}$
• M values for M-PSK and M-QAM: $M = 4$ and $M = 8$

Calculating First-Null Bandwidth

For PSK and QAM signals, the first-null bandwidth ($B$) is approximated by: $B = f_s \times (1 + \alpha)$ where:

• $f_s$ is the symbol rate,
• $\alpha$ is the roll-off factor (for ideal signals with no roll-off, $\alpha = 0$).

If no roll-off factor is specified, we assume an ideal filter ($\alpha = 0$), so: $B = f_s = 8.3 \, \text{kHz}$

Thus:

1. For 4-PSK (M=4): First-null bandwidth = 8.3 kHz
2. For 8-PSK (M=8): First-null bandwidth = 8.3 kHz
3. For 4-QAM (M=4): First-null bandwidth = 8.3 kHz
4. For 8-QAM (M=8): First-null bandwidth = 8.3 kHz

Summary

In this case, the first-null bandwidth of each signal is 8.3 kHz.

Do you have questions, or would you like further details?

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1. How does the roll-off factor impact the bandwidth in practical systems?
2. What is the difference between PSK and QAM in terms of bandwidth efficiency?
3. How would changing the symbol rate affect the first-null bandwidth?
4. Why is the first-null bandwidth the same for different M values in ideal conditions?
5. What effect does using a non-ideal filter have on bandwidth?

Tip: Higher modulation schemes (like 8-PSK or 8-QAM) provide more bits per symbol, increasing data rate at the same symbol rate, but may require better signal-to-noise ratios to maintain signal quality.